Question 511 Mark
If the function f : R → A given by $\text{f(x)}=\frac{\text{x}^2}{\text{x}^2+1}$ is a surjection, then A =
- R
- [0, 1]
- [0, 1)
- [0, 1)
Answer
- [0, 1)
Solution:
As f is surjective, range of f = co-domain of f
⇒ A = range of f
$=\frac{\text{x}^2}{\text{x}^2+1},$
$\text{y}=\frac{\text{x}^2}{\text{x}^2+1}$
$\Rightarrow\ \text{y}(\text{x}^2+1)$
$\Rightarrow\ \text{x}^2=\frac{-\text{y}}{(\text{y}-1)}$
$\Rightarrow\ \text{x}=\sqrt{\frac{\text{y}}{(1-\text{y})}}$
$\Rightarrow\ \frac{\text{y}}{(1-\text{y})}\geq0$
$\Rightarrow\ \text{y}\in[0,1)$
⇒ Range of f = [0, 1)
⇒ A = [0, 1) View full question & answer→Question 521 Mark
A relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by : xRy ⇔ x is relatively prime to y. Then, domain of R is:
- {2, 3, 5}
- {3, 5}
- {2, 3, 4}
- {2, 3, 4, 5}
Answer
- {2, 3, 4, 5}
Solution:
Given that relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by : xRy ⇔ x is relatively prime to y. R can be written as,
{(2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7)}
Here we can see that domain means x element which is $2\leq\text{x}\leq5.$
Hence, {2, 3, 4, 5} View full question & answer→MCQ 531 Mark
If a relation $R$ is defined on the set $Z$ of integers as follows: $(a, b) \in R \Leftrightarrow a^2 + b^2 = 25$. Then, domain $(R)$ is:
- A
$\{3, 4, 5\}$
- B
$\{0, 3, 4, 5\}$
- C
$\{0,\pm3,\pm4,\pm5\}$
- ✓
AnswerAs $aRb \Leftrightarrow a < b$ does not satisfy reflexive and symmetric relation.
View full question & answer→MCQ 541 Mark
The function $f : A \rightarrow B$ defined by $f(x) = -x^2 + 6x- 8$ is a bijection if,
- ✓
$\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$
- B
$\text{A}=[-3,\infty)$ and $\text{B}=(-\infty,1]$
- C
$\text{A}=(-\infty,3]$ and $\text{B}=[1,\infty)$
- D
$\text{A}=[3,\infty)$ and $\text{B}=[1,\infty)$
AnswerCorrect option: A. $\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$
$f(x) = -x^2 + 6x - 8,$ is a polynomial function and the domain of polynomial function is real number.
$\therefore\ \text{x}\in\text{R}$
$f(x) = -x^2 + 6x - 8$
$= -(x^2 - 6x + 8)$
$= -(x^2 - 6x + 9 - 1)$
$= -(x - 3)^2 + 1$
Maximum value of $-(x - 3)^2$ woud be 0
$\therefore$ Maximum value of $-(x - 3)^2 + 1$ woud be $1$
$\therefore\ \text{f(x)}\in(-\infty,1]$
We can see from the given graph that function is symmetrical about $x = 3$ and the given function is bijective.
So$, x$ would be either $(-\infty,3]$ or $[3,\infty)$
The correct option which satisfy $A$ and $B$ both is:
$\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$ View full question & answer→Question 551 Mark
Choose the correct answer from the given four options.
If the set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is:
- 720
- 120
- 0
- none of these.
Answer
- 0
Solution:
Since, the number of elements in B is more than A.
Hence, there cannot be any one-one and onto mapping from A to B.
View full question & answer→Question 561 Mark
If A = {a, b, c, d}, then a relation R = {(a, b), (b, a), (a, a)} on A is:
- Symmetric and transitive only.
- Reflexive and transitive only.
- Symmetric only.
- Transitive only.
Answer
- Symmetric and transitive only.
Solution:
Given that A = {a, b, c, d} then a relation R = {(a, b), (b, a), (a, a)} on A.
(a, b), (b, a) $\in\text{R}$
⇒ R is symmetric.
Also for (a, a) R is symmetric. View full question & answer→Question 571 Mark
If R is the largest equivalence relation on a set A and S is any relation on A, then:
- $\text{R}\subset\text{S}$
- $\text{S}\subset\text{R}$
- $\text{R = S}$
- None of these.
Answer
- $\text{S}\subset\text{R}$
Solution:
Given that R is the largest relation on A and S is any relation on A.
We know that R is always subset of A × A.
Hence, $\text{S}\subset\text{R}.$ View full question & answer→MCQ 581 Mark
$R$ is a relation from $\{11, 12, 13\}$ to ${8, 10, 12}$ defined by $y = x - 3$. Then, $R^{-1}$ is:
- ✓
$\{(8, 11), (10, 13)\}$
- B
$\{(11, 8), (13, 10)\}$
- C
$\{(10, 13), (8, 11)\}$
- D
AnswerCorrect option: A. $\{(8, 11), (10, 13)\}$
Given that $R$ is a relation from $\{11, 12, 13\}$ to $\{8, 10, 12\}$ defined by $y = x - 3$.
$R = \{(8, 11), (10, 13)\}$
$R^{-1} = \{(8, 11), (10, 13)\}$
As inverse function of $R$ is,
$y + 3 = x$
$\Rightarrow y = x + 3$
View full question & answer→Question 591 Mark
Choose the correct answer from the given four options.Let f : N → R be the function defined by $\text{f}(\text{x})=\frac{2\text{x}-1}{2}$ and g : Q → R be another function defined by g(x) = x + 2. Then $(\text{gof})\frac{3}{2}$ is:
- $1$
- $1$
- $\frac{7}{2}$
- $\text{None of these}.$
Answer
- $\text{none of these}.$
Solution:
We have $\text{f}(\text{x})=\frac{2\text{x}-1}{2}$ and g(x) = x + 2
$\text{gof}\Big(\frac{3}{2}\Big)=\text{g}\Big(\text{f}\Big(\frac{3}{2}\Big)\Big)$
$=\text{g}\bigg(\frac{2\times\frac{3}{2}-1}{2}\bigg)$
$=\text{g}(1)=1+2=3$ View full question & answer→Question 601 Mark
Let $\text{f(x)}=\frac{\alpha\text{x}}{\text{x}+1},\ \text{x}\neq-1.$ Then, for what value of $\alpha$ is f(f(x)) = x?
- $\sqrt{2}$
- $-\sqrt{2}$
- 1
- -1
Answer
- -1
Solution: Given function is $\text{f(x)}=\frac{\alpha\text{x}}{\text{x}+1},\ \text{x}\neq-1$
Also f(f(x)) = x
$\text{f}\Big(\frac{\alpha\text{x}}{\text{x}+1}\Big)=\text{x}$
$\frac{\alpha\big(\frac{\alpha\text{x}}{\text{x}+1}\big)}{\frac{\alpha\text{x}}{\text{x}+1}+1}=\text{x}$
$\frac{\alpha^2\text{x}}{\alpha\text{x}+\text{x}+1}=\text{x}$
$\alpha^2=\alpha\text{x}+\text{x}+1$
$\alpha^2=(\alpha+1)\text{x}+1$
Comparing on both sides,
$\alpha+1=0\Rightarrow\ \alpha=-1$ View full question & answer→Question 611 Mark
Let × be a binary operation on set Q of rational numbers defined as $\text{a}\times\text{b}=\frac{\text{ab}}{5}.$ Write the identity for ×.
- 5
- 3
- 1
- 6
View full question & answer→Question 621 Mark
If X is brother of the son of Y's son. How is X related to Y?
- Son
- Brother
- Cousin
- Grandson
Answer
- Grandson
Solution:
Son of Y's Son- Grandson, Brother of Y's Grandson- Y's Grandson
Option D is correct.
View full question & answer→Question 631 Mark
Let * be a binary operation on R defined by a * b = ab + 1. Then, * is:
- Commutative but not associative.
- Associative but not commutative.
- Neither commutative nor associative.
- Both commutative and associative.
Answer
- Commutative but not associative.
Solution:
Commutativity:
Let $\text{a, b}\in\text{R}$
a * b = ab + 1
= ba + 1
= b * a
Therefore,
a * b = b * a, $\forall\text{ a, b}\in\text{R}$
Therefore, * is commutative on R.
Associativity:
Let $\text{ a, b, c}\in\text{R}$
a * (b * c) = a * (bc + 1)
= a(bc + 1) + 1
= abc + a + 1
(a * b) * c = (ab + 1) * c
= (ab + 1)c + 1
= abc + c + 1
$\therefore$ a * (b * c) $\neq$ (a * b) * c
For example: a = 1, b = 2 and c = 3 [which belong to R]
Now,
1 * (2 * 3) = 1 * (6 + 1)
= 1 * 7
= 7 + 1
= 8
(1 * 2) * 3 = (2 + 1) * 3
= 3 * 3
= 9 + 1
= 10
⇒ 1 * (2 * 3) $\neq$ (1 * 2) * 3
Therefore, $\exists$ a = 1, b = 2 and c = 3 which belong to R such that
a * (b * c) $\neq$ (a * b) * c
Hence, * is not associative on R. View full question & answer→MCQ 641 Mark
Let $f(x) = x^2$ and $g(x) = 2^x$. Then, the solution set of the equation fog $(x) = gof(x)$ is:
AnswerSince $(fog)(x) = (gof)(x),$
$f(g(x)) = g(f(x))$
$\Rightarrow\ {f}(2^{x})=\text{g}({x}^2)$
$\Rightarrow\ \big(2^{\text{x}}\big)^{2}=2^{{x}^2}$
$\Rightarrow\ 2^{2{x}}=2^{{x}^2}$
$\Rightarrow\ {x}^2=2{x}$
$\Rightarrow\ {x}^2-2{x}=0$
$\Rightarrow\ {x}({x}-2)=0$
$\Rightarrow\ {x}=0, 2$
$\Rightarrow\ {x}\in\{0,2\}$
View full question & answer→Question 651 Mark
Let × be a binary operation on set Q - {1} defind by a × b = a + b - ab : a, b ∈ Q - {1}. Then × is:
- Commutative.
- Associative.
- Both (a) and (b).
- None of these.
MCQ 661 Mark
Let f : $R \rightarrow R$ be given by $\text{f(x)}=\tan {x}.$ Then, $f^{-1}(1)$ is:
AnswerCorrect option: B. $\big\{\text{n}\pi+\frac{\pi}{4}:\text{n}\in\text{Z}\big\}$
We have $, f : R \rightarrow R$ is given by
$\text{f(x)}=\tan {x}$
$\Rightarrow\ \text{f}^{-1}({x})=\tan^{-1}{x}$
$\therefore\ \text{f}^{-1}(1)=\tan^{-1}1=\big\{\text{n}\pi+\frac{\pi}{4}:\text{n}\in\text{Z}\big\}$
View full question & answer→MCQ 671 Mark
The maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ is:
AnswerThe maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ is,
$R_1 = \{(1, 1)\}$
$R_2 = \{(2, 2)\}$
$R_3 = \{(3, 3)\}$
$R_4 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$
$R_5 = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}$
The maximum number of equivalence relations on the set $A = \{1, 2, 3\}$ is $5.$
View full question & answer→Question 681 Mark
Let R be the relation over the set of all straight lines in a plane such that $\text{l}_1\text{Rl}_2\Leftrightarrow\text{l}_1\bot\text{l}_2.$ Then, R is:
- Symmetric.
- Reflexive.
- Transitive.
- An equivalence relation.
Answer
- Symmetric.
Solution:
Given R is the relation over the set of all straight lines in a plane such that $\text{l}_1\text{Rl}_2\Leftrightarrow\text{l}_1\bot\text{l}_2.$
It is symmetric relation as we can say either or $\text{l}_2\bot\text{l}_1.$ View full question & answer→MCQ 691 Mark
If the binary operation $*$ on $Z$ is defined by $a * b = a^2 − b^2 + ab + 4,$ then value of $(2 * 3) * 4$ is:
AnswerGiven that $a * b = a^2 - b^2 + ab + 4$
So,
$2 * 3$
$= 2^2 - 3^2 + 2.3 + 4$
$= 4 - 9 + 6 + 4$
$= 5$
Now,
$(2 * 3) * 4$
$= 5 * 4$
$= 5^2 - 4^2 + 5.4 + 4$
$= 25- 16 + 20 + 4$
$= 33$
View full question & answer→Question 701 Mark
An operation * is defined on the set Z of non-zero integers by a * b = ab for all a, b ∈ Z. Then the property satisfied is:
- Closure.
- Commutative.
- Associative.
- None of these.
Answer
- None of these.
Solution:
* is not clouser because when a = 1 and b = 2,
$\text{a}*\text{b}=\frac{\text{a}}{\text{b}}=\frac{1}{2}\in\text{Z}$
* is not commutative because when a = 1, b = 2 and c = 3,
$1*(2*3)=1*\Big(\frac{2}3\Big)$
$=\frac{1}{\big(\frac{2}{3}\big)}$
$=\frac{3}2$
$(1*2)*3=\frac{1}2*3$
$=\frac{\big(\frac{1}2\big)}{3}$
$=\frac{1}6$
Thus,
$1*(2*3)\neq(1*2)*3$ View full question & answer→MCQ 711 Mark
Let $f(x) = x^3$ be a function with domain ${0, 1, 2, 3}$. Then domain of $f^{-1}$ is:
- A
$\{3, 2, 1, 0\}$
- B
$\{0, -1, -2, -3\}$
- ✓
$\{0, 1, 8, 27\}$
- D
$\{0, -1, -8, -27\}$
AnswerCorrect option: C. $\{0, 1, 8, 27\}$
Given function is $f(x) = x^3$ be a function with domain $\{0, 1, 2, 3\}$.
Range $= \{0, 1^3, 2^3, 3^3\} = \{0, 1, 8, 27\}$
$f$ can be written as
$\{(0, 0), (1, 1), (2, 8), (3, 27)\}$
Hence $, f^{-1}$ can be written as
$\{(0, 0), (1, 1), (8, 2), (27, 3)\}$
Domain of $f^{-1}$ is $\{0, 1, 8, 27\}$
View full question & answer→Question 721 Mark
The smallest integer function f(x) = [x] is:
- One-one.
- Many-one.
- Both (a) and (b).
- None of these.
Question 731 Mark
The number of binary operations that can be defined on a set of 2 elements is:
- 8
- 4
- 16
- 64
Question 741 Mark
Let R be the relation on the set A = {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Then,
- R is reflexive and symmetric but not transitive.
- R is reflexive and transitive but not symmetric.
- R is symmetric and transitive but not reflexive.
- R is an equivalence relation.
Answer
- R is reflexive and transitive but not symmetric.
Solution:
Reflexivity: Clearly, $(\text{a, a})\in\text{R}\ \forall\ \text{a}\in\text{A}$
So, R is reflexive on A.
Symmetry: Since, $1,2\in\text{R},$ but $2,1\notin\text{R,}$ R is not symmetric on A.
Transitivity: Since, $1,3,3,2\in\text{R}$ and $1,2\in\text{R},$ R is transitive on A. View full question & answer→Question 751 Mark
Let R be a relation on the set N given by R = {(a, b): a = b - 2, b > 6}. Then,
- (2, 4) ∈ R
- (3, 8) ∈ R
- (6, 8) ∈ R
- (8, 7) ∈ R
Answer
- (6, 8) ∈ R
Solution:
a = b - 2 ⇒ 6 = 8 - 2 and b = 8 > 6
Hence, (6, 8) ∈ R
View full question & answer→Question 761 Mark
Consider a non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then, R is:
- Symmetric but not transitive.
- Transitive but not symmetric.
- Neither symmetric nor transitive.
- Both symmetric and transitive.
Answer
- Transitive but not symmetric.
Solution:
We have,
R = {(a, b): a is brother of b}
Let $(\text{a, b})\in\text{R}.$ Then,
a is brother of b.
but b is not necessary brother of a (As, b can be sister of a)
$\Rightarrow\ (\text{b, a})\notin\text{R}$
So, R is not symmetric.
Also,
Let $(\text{a, b})\in\text{R}$ and $(\text{b, c})\in\text{R}$
⇒ a is brother of b and b is brother of c
⇒ a is brother of c
$\Rightarrow\ (\text{a, c})\in\text{R}$
So, R is transitive. View full question & answer→MCQ 771 Mark
Let $f : R \rightarrow R$ be given by $f(x) = \tan x.$ Then $f^{-1}(1)$ is$:$
AnswerCorrect option: B. $\{\text{n}\pi+\frac{\pi}{4};\text{n }\epsilon\text{ Z}\}$
$\{\text{n}\pi+\frac{\pi}{4};\text{n }\epsilon\text{ Z}\}$
View full question & answer→Question 781 Mark
Let R be a relation on N defined by x + 2y = 8. The domain of R is:
- {2, 4, 8}
- {2, 4, 6, 8}
- {2, 4, 6}
- {1, 2, 3, 4}
Answer
- {2, 4, 6}
Solution:
The relation R is defined as R = x, y: $\text{x, y}\in\text{N}$ and x + 2y = 8
⇒ R = x, y: $\text{x, y}\in\text{N}$ and $\text{y}=\frac{8-\text{x}}{2}$
Domain of R is all values of $\text{x}\in\text{N}$ satisfying the relation R.
Also, there are only three values of x that result in y, which is a natural number.
These are {2, 6, 4}. View full question & answer→MCQ 791 Mark
$Q^+$ is the set of all positive rational numbers with the binary operation $*$ defined by $\text{a}*\text{b}=\frac{\text{ab}}2\ \forall\text{ a, b}\in\text{Q}^+$. The inverse of an element $\text{a}\in\text{Q}^+$ is:
- A
$\text{a}$
- B
$\frac{1}{\text{a}}$
- C
$\frac{2}{\text{a}}$
- ✓
$\frac{4}{\text{a}}$
AnswerCorrect option: D. $\frac{4}{\text{a}}$
Let e be the identity element in $Q^+$ with respect to $*$ such that
$a * e = a = e * a, \forall\text{ a}\in\text{Q}^+$
$a * e = a$ and $e * a = a, \forall\text{ a}\in\text{Q}^+$
$\frac{\text{ae}}2=\text{a}$ and $\frac{\text{ea}}2=\text{a}$, $\forall\text{ a}\in\text{Q}^+$
$\text{e}=2\in\text{Q}^+, \forall\text{ a}\in\text{Q}^+$
Thus $,2$ is the identity element in $Q^+$ with respect to $*$.
Let $\text{ a}\in\text{Q}^+$ and $\text{ b}\in\text{Q}^+$ be the inverse of $a$.
Then,
$a * e = a = e * a$
$a * b = e$ and $b * a = e$
$\frac{\text{ab}}2=2$ and $\frac{\text{ba}}2=2$
$\text{b}=\frac{4}{\text{a}}\in\text{Q}^+$
Thus, $\frac{4}{\text{a}}$ is the inverse of $\text{ a}\in\text{Q}^+$.
View full question & answer→Question 801 Mark
The binary operation × defined on N by a × b = a + b + ab for all a, b ∈ N is:
- Commutative only.
- Associative only.
- Both commutative and associative.
- None of these.
Answer
- Both commutative and associative.
View full question & answer→MCQ 811 Mark
Let $f : R \rightarrow R, g : R \rightarrow R$ be two functions such that $f(x) = 2x – 3, g(x) = x^3 + 5.$ The function $(fog)^{-1} (x)$ is$:$
- A
$\Big(\frac{\text{x}+7}{2}\Big)^\frac{1}{3}$
- B
$\Big(\text{x}-\frac{7}{2}\Big)^\frac{1}{3}$
- ✓
$\Big(\frac{\text{x}-2}{7}\Big)^\frac{1}{3}$
- D
$\Big(\frac{\text{x}-7}{2}\Big)^\frac{1}{3}$
AnswerCorrect option: C. $\Big(\frac{\text{x}-2}{7}\Big)^\frac{1}{3}$
$\Big(\frac{\text{x}-7}{2}\Big)^\frac{1}{3}$
View full question & answer→Question 821 Mark
The relation 'R' in N × N such that (a, b)R(c, d) ⇔ a + d = b + c is:
- Reflexive but not symmetric.
- Reflexive and transitive but not symmetric.
- An equivalence relation.
- None of the these.
Answer
- An equivalence relation.
Solution:
We observe the following properties of relation R.
Reflexivity: Let $(\text{a, b})\in\text{N}\times\text{N}$
$\Rightarrow\ \text{a, b}\in\text{N}$
$\Rightarrow\ \text{a}+\text{b}=\text{b}+\text{a}$
$\Rightarrow\ (\text{a, b})\in\text{R}$
So, R is reflexive on N × N.
Symmetry: Let $(\text{a, b}),\ (\text{c, d})\in\text{N}\times\text{N}$ such that (a, b)R(c, d)
$\Rightarrow\ \text{a}+\text{d}=\text{b}+\text{c}$
$\Rightarrow\ \text{d}+\text{a}=\text{c}+\text{b}$
$\Rightarrow\ (\text{d, c}),\ (\text{b, a})\in\text{R}$
So, R is symmetric on N × N.
Transitivity: Let $(\text{a, b}),\ (\text{c, d}),\ (\text{e, f})\in\text{N}\times\text{N}$ such that (a, b)R(c, d) and (c, d)R(e, f)
⇒ a + d = b + c and c + f = d + e
⇒ a + d + c + f = b + c + d + e
⇒ a + f = b + e
⇒ (a, b)R(e, f)
So, R is transitive on N × N.
Hence, R is an equivalence relation on N. View full question & answer→Question 831 Mark
Let f : R → R be a function defined by $\text{f(x)}=\frac{\text{e}^{|\text{x}|}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}.$ Then,
- f is a bijection.
- f is an injection only.
- f is surjection on only.
- f is neither an injection nor a surjection.
Answer
- f is neither an injection nor a surjection.
Solution:
f : R → R
$\text{f(x)}=\frac{\text{e}^{|\text{x}|}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
For x = -2 and -3 $\in\text{R}$
$\text{f(-2)}=\frac{\text{e}^{|-2|}-\text{e}^2}{\text{e}^{-2}+\text{e}^2}$
$=\frac{\text{e}^2-\text{e}^2}{\text{e}^{-2}+\text{e}^2}$
$=0$
Hence, for different values of x we are getting same values of f(x)
That means, the given function is many one.
Therefore, this function is not injective.
For x < 0
f(x) = 0
For x > 0
$\text{f(x)}=\frac{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
$=\frac{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{}e^{-\text{x}}}-\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
$=1-\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$
The value of $\frac{2\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}$ is always positive.
Therefore, the value of f(x) is always less than 1.
Numbers more than 1 are not included in the range but they are included in co-domain.
As the codomain is R.
$\therefore\ \text{Co-domain}\neq\text{Range}$
Hence, the given function is not onto.
Therefore, this function is not surjective. View full question & answer→Question 841 Mark
The law a + b = b + a is called:
- Closure law.
- Associative law.
- Commutative law.
- Distributive law.
Answer
- Commutative law.
Solution:
The law a + b = b + a is commutative.
View full question & answer→Question 851 Mark
Let A = R – {3}, B = R – {1}. Let f : A → B be defined by $\text{f(x)}=\frac{\text{x}-2}{\text{x}-3}.$ Then,
- F is bijective.
- F is one-one but not onto.
- F is onto but not one-one.
- None of these.
View full question & answer→Question 861 Mark
If the binary operation × is defind on the set Q + of all positive rational numbers by $\text{a}\times\text{b}=\frac{\text{ab}}{4}.$ Then, $3\times\Big(\frac{1}{5}\times\frac{1}{2}\Big)$ is equal to:
- $\frac{3}{160}$
- $\frac{5}{160}$
- $\frac{3}{10}$
- $\frac{3}{40}$
View full question & answer→MCQ 871 Mark
If $f : R \rightarrow R$ defined by $\text{f(x)}=\frac{3\text{x}+5}{2}$ is an invertible function, then find $f^{-1}.$
- ✓
$\frac{2\text{x}-5}{3}$
- B
$\frac{\text{x}-5}{3}$
- C
$\frac{5\text{x}-2}{3}$
- D
$\frac{\text{x}-2}{3}$
AnswerCorrect option: A. $\frac{2\text{x}-5}{3}$
$\frac{2\text{x}-5}{3}$
View full question & answer→MCQ 881 Mark
Let $f : R \rightarrow R$ be defined as $\text{f(x)}=\begin{cases}2\text{x},&\text{if x}>3\\\text{x}^2,&\text{if }1<\text{x}\leq3\\3\text{x},&\text{if x}\leq1\end{cases}.$ Then, find $f(-1) + f(2) + f(4):$
AnswerWe have,
$\text{f(x)}=\begin{cases}2\text{x},&\text{if x}>3\\\text{x}^2,&\text{if }1<\text{x}\leq3\\3\text{x},&\text{if x}\leq1\end{cases}$
Now,
$f(-1) + f(2) + f(4)$
$= 3(-1) + 2^2 + 2(4)$
$= -3 + 4 + 8$
$= 9$
View full question & answer→Question 891 Mark
Let S = {1, 2, 3, 4, 5} and let A = S × S. Define the relation R on A as follows:
(a, b) R (c, d) iff ad = cb. Then, R is;
- Reflexive only.
- Symmetric only.
- Transitive only.
- Equivalence relation.
Question 901 Mark
If A = {1, 2, 3}, then a relation R = {(2, 3)} on A is:
- Symmetric and transitive only.
- Symmetric only.
- Transitive only.
- None of these.
Answer
- Transitive only.
Solution:
The relation R is not reflexive because every element of A is not related to itself. Also, R is not symmetric since on interchanging the elements, the ordered pair in R is not contained in it.
R is transitive by default because there is only one element in it.
View full question & answer→MCQ 911 Mark
Let $f : R \rightarrow R$ be given by $f(x) = [x^2] + [x + 1] - 3$ where $[x]$ denotes the greatest integer less than or equal to $x.$ Then, $f(x)$ is:
- A
Many $-$ one and onto.
- ✓
Many $-$ one and into.
- C
One $-$ one and into.
- D
One $-$ one and onto.
AnswerCorrect option: B. Many $-$ one and into.
$f : R \rightarrow R$
$= [x^2] + [x + 1] - 3$
It is many one function because in this case for two different values of $x$ we would get the same value of $f(x).$
For $\text{x}=1.1,\ 1.2\in\text{R}$
$f(1.1) = (1.1)^2 + [1.1 + 1] - 3$
$= [1.21] + [2.1] - 3$
$= 1 + 2 + 3 = 0$
$f(1.1) = [1.2]^2 + [1.2 + 1] - 3$
$= [1.44] + [2.2] - 3$
$= 1 + 2 - 3$
$= 0$
It is into function because for the given domain we would only get the integral values of $f(x).$
But $R$ is the co $-$ domain of the given function.
That means, $\text{Co-domain}\neq\text{Range}$
Hence, the given function is into function.
Therefore, $f(x)$ is many one and into.
View full question & answer→MCQ 921 Mark
If $f : R \rightarrow R, g : R \rightarrow R$ and $h : R \rightarrow R$ is such that $f(x) = x^2, g(x) = \tan \ x$ and $h(x) = \log \ x,$ then the value of $[ho(\text{gof})](x),$ if $\text{x}=\frac{\sqrt{\pi}}{2}$ will be$:$
View full question & answer→Question 931 Mark
Choose the correct answer from the given four options.
Consider the non-empty set consisting of children in a family and a relation R defined as aRb if a is brother of b. Then R is:
- Symmetric but not transitive.
- Transitive but not symmetric.
- Neither symmetric nor transitive.
- Both symmetric and transitive.
Answer
- Transitive but not symmetric.
Solution:
We are given that a relation R defined aRb ⇒ a is brother of b.
aRa ⇒ a is brother of a, which is not true.
Hence, R is not reflexive.
aRb ⇒ a is brother of b.
This does not mean b is also a brother of a and b can be a sister of a.
Hence, it is not symmetric.
aRb ⇒ a is brother of b
and bRc ⇒ b is a brother of c.
So, a is brother of c.
Hence, R is transitive.
View full question & answer→Question 941 Mark
Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then R is:
- Reflexive but not symmetric.
- Reflexive but not transitive.
- Symmetric and transitive.
- Neither symmetric, nor transitive.
Answer
- Reflexive but not symmetric.
View full question & answer→Question 951 Mark
f : R → R is defined by $\text{f(x)}=\frac{\text{e}^{\text{x}^2}-\text{e}^{-\text{x}^2}}{\text{e}^{\text{x}^2}+\text{e}^{-\text{x}^2}}$ is:
- One-one but not onto.
- Many-one but onto.
- One-one and onto.
- Neither one-one nor onto.
Answer
- Neither one-one nor onto.
Solution:
We have,
$\text{f(x)}=\frac{\text{e}^{\text{x}^2}-\text{e}^{-\text{x}^2}}{\text{e}^{\text{x}^2}+\text{e}^{-\text{x}^2}}$
Here, $-2,2\in\text{R}$
Now, $2\neq-2$
But, f(2) = f(-2)
Therefore, function is not one-one.
And,
The minimum value of the function is 0 and maximum value is 1.
That is range of the function is [0, 1] but the co-domain of the function is given R.
Therefore, function is not onto.
$\therefore$ function is neither one-one nor onto. View full question & answer→MCQ 961 Mark
Choose the correct answer from the given four options.
Which of the following functions from $Z$ into $Z$ are bijections?
- A
$f(x) = x^3$
- ✓
$f(x) = x + 2$
- C
$f(x) = 2x + 1$
- D
$f(x) = x^2 + 1$
AnswerCorrect option: B. $f(x) = x + 2$
Consider, the second option i.e., $f(x) = x + 2$
Now, $f(x_1) = f(x_2)$
$\Rightarrow x_1 + 2 = x_2 + 2$
$\Rightarrow x_{1 }= x_2$
Hence, $f(x) = x + 2$ is one $-$ one function.
Now, let us suppose, $y = x + 2$
$\text{x}=\text{y}-2\in\text{Z},\ \forall\ \text{y}\in\text{x}$
Hence, $f(x)$ is one $-$ one and onto.
View full question & answer→Question 971 Mark
Let f : [0, $\infty$) → [0, 2] be defined by $\text{f(x)}=\frac{2\text{x}}{1+\text{x}},$ then f is:
- One-one but not onto.
- Onto but not one-one.
- Both one-one and onto.
- Neither one-one nor onto.
View full question & answer→Question 981 Mark
If R is a relation on the set A = {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3)}, then R is:
- Reflexive.
- Symmetric.
- Transitive.
- All the three options.
Answer
- All the three options.
Solution:
R = a, b : a = b and $\text{a, b}\in\text{A}$
Reflexivity: Let $\text{a}\in\text{A}$
Then, a = a
$\Rightarrow\ \text{a, a}\in\text{R}$ for all $\text{a}\in\text{A}$
So, R is reflexive on A.
Symmetry: Let $\text{a, b}\in\text{A}$ such that $\text{a, b}\in\text{R.}$
Then, $\text{a, b}\in\text{R}$
⇒ a = b ⇒ b = a ⇒ b, $\text{a}\in\text{R}$ for all $\text{a}\in\text{A}$
So, R is symmetric on A. View full question & answer→MCQ 991 Mark
If $f : R \rightarrow (-1, 1)$ is defined by $\text{f(x)}=\frac{-\text{x}|\text{x}|}{1+\text{x}^2},$ then $f^{-1}(x)$ equals,
- A
$\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$
- ✓
$-\text{Sgn (x)}\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$
- C
$-\sqrt{\frac{\text{x}}{1-\text{x}}}$
- D
AnswerCorrect option: B. $-\text{Sgn (x)}\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$
Given function is $f : R \rightarrow (-1, 1)$ is defined by $\text{f(x)}=\frac{-\text{x}|\text{x}|}{1+\text{x}^2}$
Here, for mod function we will have to consider three cases as,
$x < 0, x = 0, x > 0$
$x < 0$
$ \Rightarrow |x| = -x$
$\text{f(|x|)}=\frac{-\text{x}(-\text{x})}{1+\text{x}^2}$
$\text{y}=\frac{\text{x}^2}{1+\text{x}^2}$
$\text{y}(1+\text{x}^2)=\text{x}^2$
$\text{y}+\text{yx}^2=\text{x}^2$
$\text{y}=\text{x}^2-\text{yx}^2$
$\text{y}=(1-\text{y})\text{x}^2$
$\text{x}^2=\frac{\text{y}}{1-\text{y}}$
$\text{x}=-\sqrt{\frac{\text{y}}{1-\text{y}}}$
$\Rightarrow\ \text{x}=-\sqrt{\frac{|\text{y}|}{1-|\text{y}|}}\ \text{x} < 0$
Also you can check for the cases $x = 0$ and $x > 0$ that $\text{x}=-\sqrt{\frac{|\text{y}|}{1-|\text{y}|}}$
$-\text{Sgn (x)}\sqrt{\frac{|\text{x}|}{1-|\text{x}|}}$
View full question & answer→Question 1001 Mark
A constant function f : A → B will be one-one if:
- n(A) = n(B)
- n(A) = 1
- n(B) = 1
- n(A) < n(B)
Answer
- n(A) = 1
Solution:
Given f is a constant functions.
⇒ range of f is {c}(say)
Since f is one-one
⇒ domain of A should also contain
one element.
$\therefore\text{n(A)}=1$ View full question & answer→